In the intriguing movie *Intacto* (click here for the imdb entry for *Intacto*), luck is a more or less permanent quality of a person, provided that he or she avoids physical contact with members of a certain group who know how to steal luck. Finding people who are lucky is the quest of one of the movie's protagonists. In one case, he tests the luck of possible recruits by having them run blindfolded through a forest. The winner will get to the destination first. Many others will instead crash into the trees.

We will try a physically gentler variation on this game. In its general form, there are N people and B chances to bet. Every player knows both N and B. Each person starts with an initial wealth represented by a stack of tokens.

Each bet is an even-money bet depending on the flip of a shared fair coin. So if you bet an amount x and win, your wealth increases by x more. Otherwise you lose the x wagered.

Before each flip, each player wagers some number of tokens (from 0 to all that he or she has) on either heads or tails.

The winner is the person having the greatest number of tokens after all B bets are done. If two people tie, then nobody wins. The tokens are worthless after the betting is done. So, a player receives a reward if and only if he or she is the uncontested winner at the game's end.

In this game-like when joining a ballet company, winning the Olympics or getting ahead in a narrow hierarchy of power-the more competition there is for fewer spots, the more risk one tends to take. You'll see what I mean.

*First Warm-Up:* Suppose Bob and Alice are the only players, have the same number of tokens m and Bob must state his bet first. Suppose too there is just one remaining flip of the coin. If Alice follows the best possible strategy, what are her chances of winning?

*Solution to First Warm-Up:* If Bob bets b on heads, then Alice could bet b+1 on heads and Alice will win on heads and will lose on tails. Or Alice could bet nothing and then Alice will win on tails. Either way, Alice has a probability of 1/2 to win. The worst thing for Alice to do would be to bet b on heads, then neither she nor Bob would win.

*Second Warm-Up:* Alice and Bob are the only players again. Alice has more units than Bob. There are five more coin flips to go. If Bob states his bet first before every flip, then how can Alice maximize her chance to win?

*Solution to Second Warm-Up:* Alice can win the game every time. For each flip, Alice simply copies Bob's bet. Suppose Bob bets b on heads. Alice bets b on heads too. Whether the shared coin lands on heads or tails, Alice will end up b ahead of Bob.

**Now here are some challenges for you.**

1. Bob, Carol and Alice play. Alice has 51 tokens whereas Bob and Carol each have 50. There is one more bet to make and the betting order is Bob, Carol and then Alice. Bob and Carol collude to share the reward if either wins. How can they maximize their probability of having at least one of them win if there is just one coin flip left and they state their bets first?

2. Does the result change if Alice must state her bet first?

3. Suppose Bob has 51 tokens and Alice has 50. There are two coin flips left. Bob bets first before the penultimate flip. Alice bets first before the last flip. Bob has more than a 1/2 chance of winning. How much more?

4. Suppose Bob has 51 tokens and Alice has 50. Again, there are two coin flips left. This time, Alice bets first before the penultimate flip. Bob bets first before the last flip. Can Alice win half the time?

5. Suppose Bob has 51 tokens and Alice 50. And again, there are two coin flips left. Again, Alice states her bet first on the penultimate round and Bob states his bet first on the final round. This time, Bob announces that he will bet 20 in the penultimate round although he will wait to see Alice's bet before saying whether he will bet on heads or tails. Can Alice arrange to have more than a 1/2 chance to win?